How long does one orbit take?
The time a satellite, moon or planet takes to go once round is its orbital period, and Kepler’s third law ties it to the size of the orbit: T = 2π √(a³ ÷ GM). This orbital period calculator applies it to the Sun, any planet, the Moon or a body you define, using the semi-major axis a measured from the centre of the central body. Give an altitude above the surface (or a perigee and apogee for an elliptical orbit) and it returns the period, the orbital speed and how many orbits fit in a day. Switch the mode and it solves the other way: from a known period to the altitude needed.
It is the calculation behind satellite height questions, geostationary orbits and school problems on Kepler’s laws. Every step is shown, and everything runs in your browser.
How to use the Orbital Period Calculator
- Choose the central body, or Custom and type its mass in kilograms and radius in kilometres.
- Leave Find on “Period from orbit altitude”, enter the altitude above the surface, and optionally a higher apogee altitude for an elliptical orbit.
- Or switch Find to “Altitude from orbit period” and enter the period in minutes.
- Read the period, speeds, eccentricity and the worked steps, and compare with the table of reference orbits.
Orbital Period Calculator features
Both directions
Period from altitude, or the circular-orbit altitude that gives a chosen period.
Ellipses
Perigee and apogee give the semi-major axis, eccentricity and the vis-viva speed at each end.
Any central body
Presets for the Sun, planets, the Moon and Pluto, or your own mass and radius.
Reference orbits
Space station, imaging satellites, GPS, geostationary orbit and the Moon side by side.
Worked formula
Each substitution is written out for homework.
Formulas and a worked example
Kepler’s third law in its exact Newtonian form is T² = 4π² a³ ÷ GM. For a circular orbit the speed is v = √(GM ÷ a), and for an ellipse the speed at distance r from the centre is the vis-viva value v = √(GM (2 ÷ r − 1 ÷ a)). The semi-major axis of an ellipse is half the sum of the perigee and apogee distances, both measured from the centre, so add the body’s radius to each altitude.
Example: a satellite 400 km above Earth has a = 6,378 + 400 = 6,778 km. With GM = 3.986 × 1014 m³/s², T = 2π √((6.778 × 106)³ ÷ 3.986 × 1014) ≈ 5,554 s, about 92.6 minutes, at a speed of about 7.67 km/s. A satellite whose period equals one sidereal day (86,164 s) sits at a = 42,164 km, which is the geostationary orbit.
Altitude is not distance
The most common mistake is to put the altitude straight into the formula. The law needs the distance from the centre of the body, so a 400 km orbit around Earth uses 6,778 km, not 400 km. This page adds the radius for you, using Earth’s equatorial radius of 6,378.137 km, which is also the reference for the usual geostationary altitude of 35,786 km. If you measure altitude from a mean radius of 6,371 km instead, the same orbit reads about 7 km higher.
To know the speed needed to leave instead of orbit, use the escape velocity calculator, and for the fuel needed to reach an orbit try the rocket equation calculator. To see when the space station is overhead use the ISS pass predictor.
Limits of this calculator
- Two-body physics only: no air drag, no Earth bulge (J2), no Sun or Moon perturbations, so real satellite periods drift slightly.
- The orbiting mass is ignored; this is accurate for satellites but the Moon’s period comes out near 27.4 days instead of 27.3 because Earth’s motion about the shared centre is neglected.
- Planet radii are equatorial values from the NASA Planetary Fact Sheet; gas giants have no solid surface.
Orbital Period Calculator FAQ
What is Kepler’s third law?
The square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit. With Newton’s gravity the constant is 4π² ÷ GM.
How do you calculate the orbital period of a satellite?
Add the planet’s radius to the satellite’s altitude to get a, then use T = 2π √(a³ ÷ GM). For Earth, GM is 3.986 × 1014 m³/s².
How fast does the space station travel?
About 7.67 km/s (roughly 27,600 km/h) at 400 km altitude, completing an orbit in about 92 minutes.
What altitude has a 24-hour period?
A period of one sidereal day (23 h 56 min) corresponds to a geostationary orbit about 35,786 km above the equator.
How do I handle an elliptical orbit?
Enter the perigee and apogee altitudes. The period depends only on the semi-major axis, which is the average of the two distances from the centre.
Does the satellite mass change the period?
Not for a small satellite. Only the mass of the central body matters, so a tiny cubesat and a large space station at the same height share one period.